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Finding the general solution to the linear nonhomogeneous ODE
y''-2y'+y=(6x-2)eˣ

User Alexsmn
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Final answer:

The general solution to the given ODE involves finding the homogeneous solution, then finding a particular solution that fits the nonhomogeneous term, and combining them to form the overall general solution.

Step-by-step explanation:

Finding the general solution to the linear nonhomogeneous ordinary differential equation (ODE) y''-2y'+y=(6x-2)eˣ involves two steps. First, find the general solution of the associated homogeneous equation y''-2y'+y=0, which is typically in the form of exponentials or sinusoidals. For our case, the characteristic equation r²-2r+1=0 has roots r=1, so the homogeneous solution is yh(x) = (C1x + C2)eˣ, where C1 and C2 are constants.

Second, we find a particular solution yp to the nonhomogeneous equation. Since the nonhomogeneous term (6x-2)eˣ suggests an exponential input, we try a solution of similar form: Axeˣ for some constant A. Plugging Axeˣ into the ODE and solving for A will give us the correct particular solution. The final general solution is y(x) = yh(x) + yp(x).